Let $ABC$ be a triangle with $\angle BAC = 90^{\circ}$ and $\angle ACB = 54^{\circ}.$ We construct bisector $BD (D \in AC)$ of angle $ABC$ and consider point $E \in (BD)$ such that $DE = DC.$ Show that $BE = 2 \cdot AD.$
Source: Romania National Olympiad 2023
Tags: geometry, angles
Let $ABC$ be a triangle with $\angle BAC = 90^{\circ}$ and $\angle ACB = 54^{\circ}.$ We construct bisector $BD (D \in AC)$ of angle $ABC$ and consider point $E \in (BD)$ such that $DE = DC.$ Show that $BE = 2 \cdot AD.$